Knotting statistics for polygons in lattice tubes - birs.ca fileNicholas Beaton (Melbourne) Knotting statistics for polygons in lattice tubes March 26, 20192/21 SAPs in Z 3 Aself-avoiding

Knotting statistics for polygons in lattice tubes

Nicholas Beaton1 Jeremy Eng2 Chris Soteros2

1School of Mathematics and Statistics, University of Melbourne, Australia2Department of Mathematics and Statistics, University of Saskatchewan, Canada

The Topology of Nucleic Acids: Research at the Interface of Low-Dimensional Topology,Polymer Physics and Molecular Biology

Banff International Research StationMarch 25–29, 2019

Nicholas Beaton (Melbourne) Knotting statistics for polygons in lattice tubes March 26, 2019 1 / 21

SAPs in Z3

A self-avoiding polygon (SAP) is a simple closed walk on the edges of Z3:

Let pn be the number of n-edge polygons, defined up to translation (n must be even). Then

(p2m)m≥1 = 0, 3, 22, 207, 2412, 31754, 452640, 6840774, 108088232, 1768560270, . . .

(Known up to n = 32. [Clisby et al 2007])


SAPs in Z3



(p2m)m≥1 = 0, 3, 22, 207, 2412, 31754, 452640, 6840774, 108088232, 1768560270, . . .



SAPs in Z3



(p2m)m≥1 = 0, 3, 22, 207, 2412, 31754, 452640, 6840774, 108088232, 1768560270, . . .



Asymptotics of pn

Theorem (Hammersley 1961)

There exists κ = log µ such that

pn = exp{κn + o(n)} = eo(n)µn.

Either κ or µ (depending on who you ask) is known as the connective constant of the lattice.

For the cubic lattice [Clisby 2013]µ ≈ 4.684039931.

The eo(n) is conjectured to follow a power law, so that

pn ∼ Anα−3µn

for some constants A and α. The exponent α is expected to be universal (the same for any3-dimensional lattice), while A and µ are lattice-dependent. In 3D [Clisby & Dunweg 2016]

α ≈ 0.237209.


Asymptotics of pn







pn ∼ Anα−3µn


α ≈ 0.237209.


Asymptotics of pn







pn ∼ Anα−3µn


α ≈ 0.237209.


Knotted polygons

In three dimensions SAPs can be knotted:

Let pn(K) be the number of n-edge polygons of knot type K .


Knotted polygons cont’d

Theorem (Sumners & Whittington 1988)

There exists κ0 = log µ0 such that

pn(01) = exp{κ0n + o(n)} = eo(n)µn0.

Moreoverµ0 < µ.

That is, the probability of a random n-edge polygon being unknotted decays exponentially:

Pn(01) =pn(01)

pn= eo(n)

(µ0

µ

)n

.

But the decay is still slow [Janse van Rensburg 2008]:

µ0

µ≈ 0.999996.



Theorem (Sumners & Whittington 1988)

There exists κ0 = log µ0 such that

pn(01) = exp{κ0n + o(n)} = eo(n)µn0.

Moreoverµ0 < µ.

That is, the probability of a random n-edge polygon being unknotted decays exponentially:

Pn(01) =pn(01)

pn= eo(n)

(µ0

µ

)n

.

But the decay is still slow [Janse van Rensburg 2008]:

µ0

µ≈ 0.999996.



For K 6= 01 very little has been proved. It is straightforward to show that

lim supn→∞

1

nlog pn(K) ≥ κ0.

But it is generally believed that

limn→∞

1

nlog pn(K) = κ0,

and moreoverpn(K) ∼ AKn

α−3+f (K)µn0

where α is the same exponent as for all polygons, AK is a constant and f (K) is the number ofprime knot factors in the knot decomposition of K .

This is essentially because the “knotted parts” of a long polygon are expected to be small:




lim supn→∞

1

nlog pn(K) ≥ κ0.


limn→∞

1

nlog pn(K) = κ0,


α−3+f (K)µn0






lim supn→∞

1

nlog pn(K) ≥ κ0.


limn→∞

1

nlog pn(K) = κ0,


α−3+f (K)µn0




Polygons in lattice tubes

Let TL,M ≡ T be the L×M infinite tube of Z3:

TL,M = {(x , y , z) | 0 ≤ y ≤ L, 0 ≤ z ≤ M}

and let pT,n be the number of SAPs in T, defined up to translation in the x-direction only.

z

x

y

In T, polygons are characterised by a finite transfer matrix ⇒ growth rates, exponents, etc. canbe computed exactly (in theory).

Theorem (Soteros 1998)

There exist constants AT and κT = log µT such that

pT,n ∼ ATµnT

AT and µT are algebraic numbers.


Polygons in lattice tubes

Let TL,M ≡ T be the L×M infinite tube of Z3:

TL,M = {(x , y , z) | 0 ≤ y ≤ L, 0 ≤ z ≤ M}

and let pT,n be the number of SAPs in T, defined up to translation in the x-direction only.

z

x

y

In T, polygons are characterised by a finite transfer matrix ⇒ growth rates, exponents, etc. canbe computed exactly (in theory).


There exist constants AT and κT = log µT such that

pT,n ∼ ATµnT

AT and µT are algebraic numbers.


Knotted polygons in tubes

If L,M > 0 and (L,M) 6= (1, 1) then polygons in T can be knotted. (Only unknots in 1× 1.)Define pT,n(K) to count polygons of knot type K .

Similarly to Z3:


There exists κT,0 = log µT,0 such that

pT,n(01) = exp{κT,0n + o(n)} = eo(n)µnT,0.

If L,M > 0 and (L,M) 6= (1, 1) thenµT,0 < µT.

That is, in a 3-dimensional tube other than 1× 1, the probably of a polygon being unknotteddecays exponentially.


Knotted polygons in tubes cont’d

As with Z3, it is easy to show that

lim supn→∞

1

nlog pT,n(K) ≥ κT,0.

And it is believed thatpT,n(K) ∼ AT,Kn

f (K)µnT,0.

This is again because the “knotted parts” of a long polygon are expected to be small:




lim supn→∞

1



f (K)µnT,0.





lim supn→∞

1



f (K)µnT,0.



Counting by span

The x-span of a polygon is the maximal difference in x-coordinates between any two vertices.

In T, we can count polygons by x-span instead of length: qT,s and qT,s(K). The same theoremshold:

Theorem (Atapour 2008)

There exist constants BT and χT = log νT such that

qT,s ∼ BTνsT

and constants χT,0 = log νT,0 such that

qT,s(01) = exp{χT,0s + o(s)} = eo(s)νsT,0.

If L,M > 0 and (L,M) 6= (1, 1) thenνT,0 < νT.

Again expect thatqT,s(K) ∼ AT,K s

f (K)νsT,0.

BT and νT are algebraic numbers.


Counting by span





qT,s ∼ BTνsT





f (K)νsT,0.



Counting by span





qT,s ∼ BTνsT





f (K)νsT,0.



Computing probabilities

Polygons in T are characterised by a finite transfer matrix, but polygons of fixed knot type(including unknots) are not. This is because the state depends on the previous entanglement ofthe “strands”, and there are infinitely many possibilities:

In general can only estimate the behaviour of pT,n(K) and qT,s(K).

Analysing series data is hopeless because knots are not common until very big lengths / spans.

Must use Monte Carlo methods!




















Transfer matrix Monte Carlo methodA 1-pattern is any configuration of vertices and (half)-edges between x = k ± 1

2which can form

part of a polygon, together with a pairing of the open half-edges on the left.

A 1-pattern occurring at the very left end of a polygon is a starting 1-pattern (set A); at the veryright is an ending 1-pattern (set B); otherwise it is internal (set I).

Let M be the transfer matrix for internal 1-patterns, ie. Mij = 1 if i can be immediately followedby j and 0 otherwise. Since M is irreducible and aperiodic, it has a unique dominant eigenvalueλ ∈ R>0 with right eigenvector ζ.

Lemma (adapted from Alm & Janson 1990)

If i and j are internal 1-patterns such that j can follow i , let pspij (s) be the probability that an

occurrence of i in a uniformly random polygon of span s is followed by j . Then as s →∞,

pspij (s)→ psp

ij = λ−1 ζj

ζi.



2which can form







pspij (s)→ psp

ij = λ−1 ζj

ζi.



2which can form







pspij (s)→ psp

ij = λ−1 ζj

ζi.



2which can form







pspij (s)→ psp

ij = λ−1 ζj

ζi.


Transfer matrix Monte Carlo method cont’dCan likewise define matrices A with rows indexed by A and columns indexed by I, and B indexedby I and B.

We then use an algorithm for generating a polygon π = π0π1 · · ·πs uniformly at random. Fora ∈ A and i ∈ I, define

Fstart(a) = {j ∈ I : Aaj = 1} and Fend(i) = {b ∈ B : Bib = 1}

and then

t1(a) =∑

j∈Fstart(a)

ζj and ts(i) =|Fend(i)|

ζi

1 Select π0 uniformly at random from S.

2 With probability r1(π0) (see below) reject the sample and start from (1) again. Else select π1

from Fstart(π0) with probability proportional to ζπ1 .

3 For k = 2, 3, . . . , s − 1, choose πk with probability pspπk−1,πk

.

4 With probability rs(πs−1) (see below), reject the sample and start from (1) again. Otherwiseselect πs uniformly from Fend(πs−1).

The rejection probabilities are chosen to make the sampling uniformly random:

r1(π0) = 1−t1(π0)

maxa∈S{t1(a)}

and rs(πs−1) = 1−ts(πs−1)

maxj∈I{ts(j)}


Transfer matrix Monte Carlo method cont’dCan likewise define matrices A with rows indexed by A and columns indexed by I, and B indexedby I and B.

We then use an algorithm for generating a polygon π = π0π1 · · ·πs uniformly at random. Fora ∈ A and i ∈ I, define

Fstart(a) = {j ∈ I : Aaj = 1} and Fend(i) = {b ∈ B : Bib = 1}

and then

t1(a) =∑

j∈Fstart(a)

ζj and ts(i) =|Fend(i)|

ζi

1 Select π0 uniformly at random from S.

2 With probability r1(π0) (see below) reject the sample and start from (1) again. Else select π1

from Fstart(π0) with probability proportional to ζπ1 .

3 For k = 2, 3, . . . , s − 1, choose πk with probability pspπk−1,πk

.

4 With probability rs(πs−1) (see below), reject the sample and start from (1) again. Otherwiseselect πs uniformly from Fend(πs−1).

The rejection probabilities are chosen to make the sampling uniformly random:

r1(π0) = 1−t1(π0)

maxa∈S{t1(a)}

and rs(πs−1) = 1−ts(πs−1)

maxj∈I{ts(j)}


Computing probabilities cont’d

For polygons of thousands of edges, determining knot type directly is very difficult!

But we can slice them up at “2-sections” – x-coordinates with only two edges crossing – andclose up the pieces to form a sequence of smaller polygons.

Then we compute the knot type of each, and the overall knot type is the connect-sum of theparts. Since long polygons have a positive density of 2-sections (guaranteed by patterntheorems), the pieces are almost always very small.












Fixed-length vs. fixed-span vs. Hamiltonian

The transfer matrix Monte Carlo method described above was for fixed-span s, but it can beadapted to fixed-length as well.

In T it may seem that the choice is arbitrary, but in fact there is a big difference.

Fixed-span polygons are much denser. On average (edges per unit span):

fixed-length fixed-span2× 1 3.6214 4.88653× 1 4.1052 6.5244

Denser ⇒ more “entanglement” ⇒ higher knotting probability. Here I will focus on fixed-span.

The method also works for Hamiltonian polygons, which are as dense as possible (6 and 8 edgesper unit span respectively), so we sample those too. (All the previous theorems aboutasymptotics also apply.)

Define

P(sp)T,s (K) = probability of knot-type K among all knots of span s in T

PHT,s(K) = probability of knot-type K among all Hamiltonian knots of span s in T

P∗T,s(K) = one of the above









Define












Define












Define












Define












Define





Results: Growth ratesSince P∗T,s(01) decays exponentially, plot the log and take a linear best fit.

2× 1 tube:

−0.0014

−0.0012

−0.001

−0.0008

−0.0006

−0.0004

−0.0002

0

0.0002

0 200 400 600 800 1000 1200 1400

log P(sp)T,s (01)

(−8.12115× 10−7)s + 6.08976× 10−6

SoχT,0 − χT = −8.12115× 10−7 ⇒

νT,0

νT= 0.99999919



2× 1 tube:

−0.0014

−0.0012

−0.001

−0.0008

−0.0006

−0.0004

−0.0002

0

0.0002

0 200 400 600 800 1000 1200 1400

log P(sp)T,s (01)

(−8.12115× 10−7)s + 6.08976× 10−6

SoχT,0 − χT = −8.12115× 10−7 ⇒

νT,0

νT= 0.99999919



3× 1 tube:

−1.2

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0 1000 2000 3000 4000 5000 6000 7000 8000

log P(sp)T,s (01)

−0.000140089s + 0.00066107

SoχT,0 − χT = −0.000140089 ⇒

νT,0

νT= 0.99986



2× 1 tube Hamiltonian:

−0.045

−0.04

−0.035

−0.03

−0.025

−0.02

−0.015

−0.01

−0.005

0

0.005

0 200 400 600 800 1000 1200 1400

log PHT,s(01)

(−2.86675× 10−5)s + 0.000179821

So

χHT,0 − χ

HT = −2.86675× 10−5 ⇒

νHT,0

νHT= 0.999971




−4

−3.5

−3

−2.5

−2

−1.5

−1

−0.5

0

0.5

0 1000 2000 3000 4000 5000

log PHT,s(01)

−0.00713521s + 0.00277423

So

χHT,0 − χ

HT = −0.00713521 ⇒

νHT,0

νHT= 0.99289


Results: ExponentsNo evidence of logarithmic corrections: strongly suggests that

qT,s(01) ∼ BT,0(νT,0)s and qHT,s(01) ∼ BHT,0(νHT,0)s .

For other knot types, examine ratio P∗T,s(K)/P∗T,s(01).

2× 1 tube:

0

0.0002

0.0004

0.0006

0.0008

0.001

0.0012

0.0014

0 200 400 600 800 1000 1200 1400

P(sp)T,s (31)

P(sp)T,s (01)

’





2× 1 tube:

0

0.0002

0.0004

0.0006

0.0008

0.001

0.0012

0.0014

0 200 400 600 800 1000 1200 1400

P(sp)T,s (31)

P(sp)T,s (01)

’





2× 1 tube:

0

0.0002

0.0004

0.0006

0.0008

0.001

0.0012

0.0014

0 200 400 600 800 1000 1200 1400

P(sp)T,s (31)

P(sp)T,s (01)

’





3× 1 tube:

0

0.2

0.4

0.6

0.8

1

1.2

0 1000 2000 3000 4000 5000 6000 7000 8000

P(sp)T,s (31)

P(sp)T,s (01)

125P(sp)T,s (41)

P(sp)T,s (01)

1250P(sp)T,s (51)

P(sp)T,s (01)

2200P(sp)T,s (52)

P(sp)T,s (01)






0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

0.045

0 200 400 600 800 1000 1200 1400

PHT,s (31)

PHT,s (01)

285PHT,s (41)

PHT,s (01)






0

0.5

1

1.5

2

2.5

3

3.5

4

0 1000 2000 3000 4000 5000

PHT,s (31)

PHT,s (01)

50PHT,s (41)

PHT,s (01)

245PHT,s (51)

PHT,s (01)

380PHT,s (52)

PHT,s (01)


Results: Exponents cont’dFor composite knots, take the ratio P∗T,s(K)/P∗T,s(01) and then look at log-log plot.

3× 1 tube:

−20

−18

−16

−14

−12

−10

−8

−6

−4

−2

0

2

4 5 6 7 8 9

P(sp)T,s (1fact)/P(sp)

T,s (01)


T,s (01)


T,s (01)

1.00012 log s − 8.87461

2.00284 log s − 18.4666

3.02684 log s − 28.6496

Altogether, strongly implies

qT,s(K) ∼ BT,K sf (K)(νT,0)s and qHT,s(K) ∼ BH

T,K sf (K)(νHT,0)s .



3× 1 tube:

−20

−18

−16

−14

−12

−10

−8

−6

−4

−2

0

2

4 5 6 7 8 9


T,s (01)


T,s (01)


T,s (01)

1.00012 log s − 8.87461

2.00284 log s − 18.4666

3.02684 log s − 28.6496







−14

−12

−10

−8

−6

−4

−2

0

2

4

4 5 6 7 8

PHT,s(1fact)/PH

T,s(01)

PHT,s(2fact)/PH

T,s(01)

PHT,s(3fact)/PH

T,s(01)

1.00002 log s − 7.24373

2.00865 log s − 15.2526

3.02006 log s − 23.6931







−14

−12

−10

−8

−6

−4

−2

0

2

4

4 5 6 7 8

PHT,s(1fact)/PH

T,s(01)

PHT,s(2fact)/PH

T,s(01)

PHT,s(3fact)/PH

T,s(01)

1.00002 log s − 7.24373

2.00865 log s − 15.2526

3.02006 log s − 23.6931





Results: Probability maxima

P∗T,s(K) decays exponentially for any fixed knot type K or set of knot types. But if K 6= 01 then

P∗T,s(K) initially increases, reaches some maximum, then decreases to 0.

If

P∗T,s(K) ∼ C∗T,K sf (K)

(ν∗T,0

ν∗T

)s

then the maximum should be at roughly

s∗ ≈ M∗T(K) =f (K)

χ∗T − χ∗T,0

=

(1.23× 106)f (K) 2× 1 tube

7140f (K) 3× 1 tube

34900f (K) 2× 1 tube Hamiltonian



Results: Probability maxima

P∗T,s(K) decays exponentially for any fixed knot type K or set of knot types. But if K 6= 01 then

P∗T,s(K) initially increases, reaches some maximum, then decreases to 0.

If

P∗T,s(K) ∼ C∗T,K sf (K)

(ν∗T,0

ν∗T

)s

then the maximum should be at roughly

s∗ ≈ M∗T(K) =f (K)

χ∗T − χ∗T,0

=

(1.23× 106)f (K) 2× 1 tube

7140f (K) 3× 1 tube




Results: Probability maxima cont’d

Prime knots in the 3× 1 tube:

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0 1000 2000 3000 4000 5000 6000 7000 8000

P(sp)T,s (31)

123P(sp)T,s (41)

1250P(sp)T,s (51)

2150P(sp)T,s (52)



Prime Hamiltonian knots in the 3× 1 tube:

−0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0 1000 2000 3000 4000 5000

PHT,s(31)

50.5PHT,s(41)

248PHT,s(51)

380PHT,s(52)

5500PHT,s(61)

8800PHT,s(62)

50000PHT,s(63)



2-factor knots in the 3× 1 tube:

−0.05

0

0.05

0.1

0.15

0.2

0.25

0.3

0 1000 2000 3000 4000 5000 6000 7000 8000

P(sp)T,s (31#31)

61P(sp)T,s (31#41)

14500P(sp)T,s (41#41)



2-factor Hamiltonian knots in the 3× 1 tube:

−0.05

0

0.05

0.1

0.15

0.2

0.25

0.3

0 1000 2000 3000 4000 5000

PHT,s(31#31)

25PHT,s(31#41)

2600PHT,s(41#41)



Multi-factor knots in the 3× 1 tube:

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0 1000 2000 3000 4000 5000 6000 7000 8000

P(sp)T,s (1fact)

P(sp)T,s (2fact)

P(sp)T,s (3fact)



Multi-factor Hamiltonian knots in the 3× 1 tube:

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0 1000 2000 3000 4000 5000

PHT,s(1fact)

PHT,s(2fact)

PHT,s(3fact)


Closing commentsBelieve all the same to be true for fixed-length, but knots are very very rare in narrow tubes.

Transfer matrix method very memory intensive. Can (just barely) do Hamiltonian polygonsin 2× 2 and 4× 1. For bigger tubes, other methods like PERM, Wang-Landau, etc may beneeded.

In the bulk, numerical evidence [Janse van Rensburg & Rechnitzer 2011] that the amplituderatios AK1

/AK2are universal. Does not appear to be the case in tubes.

In 2× 1 only, a stronger result can be proved [Atapour, NRB, Eng, Ishihara, Shimokawa,Soteros & Vazquez, in preparation]:

pT,n(K) ∼ DT,Knf (K)pn,T(01)

if all of K ’s factors have unknotting number one.

NRB, Eng & SoterosKnotting statistics for polygons in lattice tubesJ. Phys. A: Math. Theor. 52 (2019), 144003.

Thank you!










Thank you!










Thank you!










Thank you!










Thank you!


Documents

Knotting statistics for polygons in lattice tubes - birs.ca fileNicholas Beaton (Melbourne) Knotting statistics for polygons in lattice tubes March 26, 20192/21 SAPs in Z 3 Aself-avoiding